The general solution of sin x + cos x = 1 is | vedclass

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Question

(C) Given equation: $\sin x + \cos x = 1$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}

Vedclass · Accepted Answer

$x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}, n \in Z$

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(C) Given equation: $\sin x + \cos x = 1$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$ $\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $\sin(x + \frac{\pi}{4}) = \sin \frac{\pi}{4}$ The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$. Therefore,$x + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$ $x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}, n \in Z$.

The general solution of $\sin x + \cos x = 1$ is

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